In the scope of my scheduling research, the question has been raised on whether distance-preserving permutations can be constructed easily.

Suppose that our domain is the set of natural numbers up until $n$, that is: $D=\{1,2,...,n\}$ where $n\geq 1$.

A permutation $\pi$ is a distance-preserving permutation in my context is one that has the following property: $|\pi(i) - \pi(j)| = |i-j|$.

For example, the most trivial distance-preserving permutation I have come up with is the following: $\pi(i) = n-i+1$, which is the permutation that maps $1$ to $n$, $2$ to $n-1$, etc. It can be easily shown that $|\pi(i) - \pi(j)| = |i-j|$.

My intuition says that there are no more distance-preserving permutations, since if we draw this as a graph where the $x$-axis is the numbers in $D$ and the $y$-axis is the values $\pi(i)$ then we will have a $90$-degree cornered triangle with the corner facing north-east, its' equal edges being the distances $|i-j|$ and $|\pi(i) - \pi(j)|$.

Are there distance-preserving permutations other than the above trivial one?


1 Answer 1


The only permutations who satisfy this condition are the identity and its inverse (as string inverse), $\pi(i)=n-i+1$.

Note that $1$ has to be in one of the edges (if it has two neighbors, then at least one of them is bigger than $2$ and the condition is not satisfied). Placing $1$ in the left edge forces the entire permutation to be the identity, and placing $1$ in the right edge forces the permutation to be its inverse.


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